m(m+2)+(m+3)=23

Solution for m(m+2)+(m+3)=23 equation:

m(m+2)+(m+3)=23

We move all terms to the left:

m(m+2)+(m+3)-(23)=0

We multiply parentheses

m^2+2m+(m+3)-23=0

We get rid of parentheses

m^2+2m+m+3-23=0

We add all the numbers together, and all the variables

m^2+3m-20=0

a = 1; b = 3; c = -20;
Δ = b2-4ac
Δ = 32-4·1·(-20)
Δ = 89
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{89}}{2*1}=\frac{-3-\sqrt{89}}{2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{89}}{2*1}=\frac{-3+\sqrt{89}}{2}
$